\(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\)

\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)

\(=\frac{1}{5}-\frac{1}{50}=\frac{10}{50}-\frac{1}{50}=\frac{9}{50}\)

\(=\frac{10-5}{5.10}+\frac{15-10}{10.15}+...+\frac{50-45}{45.50}\)

\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)

\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)

\(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\)

\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)

\(=\frac{1}{5}-\frac{1}{50}=\frac{10}{50}-\frac{1}{50}\)

\(=\frac{9}{50}\)

~ Rất vui vì giúp đc bn ~

\(a=3\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{45.50}\right)\)

\(a=\frac{3}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\right)\)

\(a=\frac{3}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\right)\)

\(a=\frac{3}{5}.\left(\frac{1}{5}-\frac{1}{50}\right)\)

\(a=\frac{3}{5}\cdot\frac{9}{50}\)

\(a=\frac{27}{250}\)

=(5/5-5/10+5/10-5/15+.........+5/2015-5/2020)

=(1/5-1/10+1/10-1/20+.......+1/2015-1/2020)

=1/5-1/2020

=403/2020

ai tích mk mk vs

\(\frac{5}{5.10}+\frac{5}{10.15}+.............+\frac{5}{2015.2020}\)

\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..............+\frac{1}{2015}-\frac{1}{2020}\)

\(=\frac{1}{5}-\frac{1}{2020}\)

\(=\frac{403}{2020}\)

5/5.10+5/10.15+5/15.20+...+5/2015.2020

=5(1/5.10+1/10.15+1/15.20+...+1/2015.2020) khoang cach tu 5-10;10-15;...;2015-2020 la 5 suy ra

=5/5(1/5-1/10+1/10-1/15+1/15-1/20+...+1/2015-1/2020)    ;   (-1/5+1/5;-1/10+1/10;-1/15+1/15;-1/20+1/20;... bang 0)

=1(1/5-1/2020)=2015/10100=403/2020

\(B=\frac{5}{5\cdot10}+\frac{5}{10\cdot15}+...+\frac{5}{95\cdot100}\)

\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

\(B=\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)

\(B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)

\(B=\frac{1}{5}-\frac{1}{100}\)

\(B=\frac{19}{100}\)

b=\(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)

b=\(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{`1}{100}\)

b=\(\frac{1}{5}-\frac{1}{100}\)

b=\(\frac{19}{100}\)

 hoc tot nha bn !

\(P=\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{95.100}\)

\(\Rightarrow P=\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\)

\(\Rightarrow P=\dfrac{1}{5}-\dfrac{1}{100}\)

\(\Rightarrow P=\dfrac{19}{100}\)

Vậy \(P=\dfrac{19}{100}\)

a)          ta có công thức \(\frac{a}{n.\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)

ta có \(N=\frac{5^2}{5.10}+\frac{5^2}{10.15}+...+\frac{5^2}{2005.2010}\)

\(N=5\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2005.2010}\right)\)

 \(N=5\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)(sử dụng quy tắc dấu ngoặc)

\(N=5\left[\frac{1}{5}-\left(\frac{1}{10}-\frac{1}{10}\right)-\left(\frac{1}{15}-\frac{1}{15}\right)-...-\left(\frac{1}{2005}-\frac{1}{2005}\right)-\frac{1}{2010}\right]\)

\(N=5\left[\frac{1}{5}-0-0-...-0-\frac{1}{2010}\right]\)

\(N=5\left[\frac{1}{5}-\frac{1}{2010}\right]\)

\(N=5.\frac{401}{2010}\)

\(N=\frac{401}{402}\)

b)         \(M=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)

               ta thấy      \(\frac{1}{11}=\frac{1}{11}\)

                                \(\frac{1}{12}